Theory:

A solution of an equation is a number substituted for an unknown variable which makes the equality in the equation true.
In a two-variable equation, the solution also contains two values.
Method I
Consider an equation with two variable \(x + 2y = 20\).
 
We need to substitute values for \(x\) and \(y\) in \(LHS\), which should satisfy the equation and give the result value as \(RHS\).
 
Let us substitute \(x = 2\) and \(y = 9\).
 
\(LHS = x + 2y\)
 
\(= 2 + 2(9)\)
 
\(= 20 = RHS\)
 
So, \(x = 2\) and \(y = 9\) satisfy the given equation, which is the solution of the given equation.
 
Therefore, the solution can be written as \((2, 9)\).
Is there only one solution available for the single linear equation in two variables?
 
No, we can find many solutions for the single linear equation in two variables.
Substitute \(x = 10\) and \(y = 5\).
 
\(LHS = x + 2y\)
 
\(= 10 + 2(5)\)
 
\(= 20 = RHS\)
 
So, \((10, 5)\) is also a solution to the given equation.
 
Important!
We can find many solutions for a single equation with two variables.
Method II
Steps to find a solution of equation:
1. Substitute \(x = 0\) and \(y = 0\) in the given equation.
 
2. Next, perform an arithmetic operation to find the value of \(x\) and \(y\), which are the solutions of  the given equation.
Consider the same equation: \(x + 2y = 20\)
 
Substitute \(x = 0\) in the given equation.
 
\(0 + 2y = 20\)
 
\(2y = 20\)
 
y=202
 
\(y = 10\)
 
Thus, \((0, 10)\) is one solution of the given equation.
 
Similarly, substitute \(y = 0\) in the given equation.
 
\(x + 2(0) = 20\)
 
\(x + 0 = 20\)
 
\(x = 20\)
 
Thus, \((20, 0)\) is another solution to the given equation.
 
Important!
The solution of a linear equation is not affected when:
 
(i) the same number is added to (or subtracted from) both sides of the equation.
 
(ii) you multiply or divide both sides of the equation by the same non-zero number.